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The B.U.D Technique

Jay Cruz
Hey there! I'm Jay. I love to learn and am super passionate about my field, Software Engineering.
Originally published at jay-cruz.Medium on ・4 min read

Optimizing your algorithms is an important skill for new Software Engineers to develop, especially when it comes to technical interviewing. It’s a skill I’m still very much trying to learn and progress at. Lately, I’ve been reading the book Cracking the Coding Interview by Gayle Laakmann McDowell. I came across some useful techniques that are mentioned by the author. One such technique that caught my eye was B.U.D which stands for:

• B ottlenecks
• U nnecessary Work
• D uplicated Work

The basic gist of BUD is it represents the three most common tasks that an inefficient solution is wasting time with. An approach would be to walk through you’re brute force algorithm checking for each of these. Then focus on fixing it until it can be improved. So let’s break down the specifics of each of these and explain how identifying them in your algorithm using the BUD technique can help to optimize your solution.

Bottlenecks

The bottleneck is the part of your algorithm that considerably slows down its runtime. For an example say we have the following problem to solve:

Problem : You have two arrays with distinct elements in common. How would you compute the number of elements in common.

Brute force approach : Take each element in array A, walk it through array B and check if it contains the current element.

This would roughly add up to an O(A * B) runtime. The bottleneck, in this case, is B. We’re checking for each element of A to see if it contains any of the elements in B, the contains operation is O(B). This sort of repetition negatively affects our runtime. After recognizing this bottleneck we should think about what’s a more optimal way to perform this same type of operation. How about a hash lookup.

Optimizing for a bottleneck : Add every element in array B into a hash table. Then for each element in A check if it exists in the hash table.

This would bring our runtime down to O(A + B), getting rid of the bottleneck.

Unnecessary Work

For unnecessary work, we’ll basically restructure the algorithm by finding the other unneeded operations being done and make small changes to adjust for better optimization. Let’s take the following example problem from the Cracking the Coding Interview book:

Problem : Print all positive integer solutions to the equation a³ + b³ = c³ + d³ where a, b, c and d are integers between 1 and 1000. — Cracking the Coding Interview

Brute force approach : Just use several nested for loops like the following Javascript example:

``````for (let a = 1; 1 < 1000; a++) {
for (let b = 1; 1 < 1000; b++) {
for (let c = 1; 1 < 1000; c++) {
for (let d = 1; d < 1000; d++) {
if (Math.pow(a, 3) + Math.pow(b, 3) ==
Math.pow(c, 3) + Math.pow(d, 3)) {
console.log(a, b, c, d)
}
}
}
}
}
// Note: I don't suggest running this code in you're console
``````

This code solves our problem but it uses an unnecessary amount of work. To optimize we could break from the loop as soon as we find our solution. In this case, we’ll break out of the inner loop of d.

Optimize for unnecessary work :

``````for (let a = 1; 1 < 1000; a++) {
for (let b = 1; 1 < 1000; b++) {
for (let c = 1; 1 < 1000; c++) {
for (let d = 1; d < 1000; d++) {
if (Math.pow(a, 3) + Math.pow(b, 3) ==
Math.pow(c, 3) + Math.pow(d, 3)) {
console.log(a, b, c, d)
break // solution found, break from loop
}
}
}
}
}
``````

This may be a very small optimization that does little to improve our code but if you continue to make small adjustments like this it may be able to affect the overall runtime of you’re algorithm.

Duplicated Work

Here we’ll look for any duplicated work using the same example as before.

Problem : Print all positive integer solutions to the equation a³ + b³ = c³ + d³ where a, b, c and d are integers between 1 and 1000. — Cracking the Coding Interview

Optimize for duplicated work : Once again this is a problem where a hash table can be used to optimize. Add all c, d pairs to an array. Then store it as the value in the hash map.

``````const map = {}
for (let c = 1; c < 1000; c++) {
for (let d = 1; d < 1000; d++) {
let result = Math.pow(c, 3) + Math.pow(d, 3)
map[result] = [c, d]
}
}

for (let a = 1; a < 1000; a++) {
for (let b = 1; b < 1000; b++) {
let result = Math.pow(a, 3) + Math.pow(b, 3)
let list = map[result]
for (const pair of list) {
console.log(a, b, pair)
}
}
}
``````

We realized that we were duplicating our work by performing multiple nested for loop iterations. So to solve this we split up our work into two separated nested loops. The first nested loop calculates the result of c³ + d³ then mapping this to the c,d pairs array in a hash map. The second nested loop does the same calculations but for a, b pairs then we’re simply printing the results. This would bring our runtime down to O(n²) from the previous O(n⁴).

Resources

Originally published at https://coderjay06.github.io on June 26, 2021.